Finding the equation of the tangent

[Prostagma]

Hi, just a quick question as to what I'm doing wrong here. The question says to find the equation of the tangent at point (1,9) on the graph of f(x)=5x^2-6x+10

So I have

5x^2-6x+10-9
----------------
x-1

=5x^2-6x+1
--------------
x-1

=(5x-1)(x-1)
--------------
(x-1)

=(5x-1)

I then plug in 1.001 and 0.999 into the equation to find out the slope of the tangent

[5(1.001)-1]-[5(1)-1)]
-------------------------
1.001-1

=5
AND from the left

[5(0.999)-1]-[5(1)-1)]
-------------------------
0.999-1

=5

Therefore, I know that the slope of the tangent = 5

So I plug it into y=mx+b

I get 9=5(1)+b

9=5+b
9-5=b
4=b

And the equation of the tangent is y=5x+4 but the back of the book says the answer is y=4x+5

Just wanted to get some clarification please

 

[tkhunny]

There is something seriously wrong with this methodology. Why would the behavior of the quadratic be particularly related to the linear function so created? It makes no sense. Are you SURE you are not mixing up two things?

My best evidence is the rather silly calculation of the slope from two sides. It's a line with slope 5. Why would any such calculation produce anything but 5?

On the other hand, if you just forget the division problem you started with and do the left-right thing with the original function, you get slopes of 4.005 and 3.995. These are much more reasonable results.

Where have you encountered such a method?

 

[Prostagma]

I don't know :?

Our teacher told us to do it this way. I.e, find the slope of the tangent, then plug the slope of the tangent into the y=mx+b formula to solve for b, and then get the equation of the tangent.

mmm but yes the way you're saying makes sense.

if you get a slope of 4

9=4(1)+b
9-4=b

so the equation of the tangent would be y=4x+5